Thread: Series solution for y"+x(e^-x) y = 0

1. Series solution for y"+x(e^-x) y = 0

Anyone can help me ?

Question: Find series solution about the origin for
y"+x(e^-x) y = 0
and determine the lower bound for the radius of convergence.

TQ.

2. Originally Posted by younhock
Anyone can help me ?

Question: Find series solution about the origin for
y"+x(e^-x) y = 0
and determine the lower bound for the radius of convergence.

TQ.
Presumably you know that $\displaystyle e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$ so that $\displaystyle xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$.

Let $\displaystyle y= \sum_{n=0}^\infty a_nx^n$ so that $\displaystyle y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2}$ and put that into the differential equation.

3. Originally Posted by HallsofIvy
Presumably you know that $\displaystyle e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$ so that $\displaystyle xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$.

Let $\displaystyle y= \sum_{n=0}^\infty a_nx^n$ so that $\displaystyle y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2}$ and put that into the differential equation.

But i do not know how to simplify the [ $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$ ] [ $\displaystyle \sum_{n=0}^\infty a_nx^n$ ] .

thanks for guiding...